# Solving Leetcode Interviews in Seconds with AI: Count Distinct Numbers on Board


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2549" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure:  For each number x present on the board, find all numbers 1 <= i <= n such that x % i == 1. Then, place those numbers on the board.  Return the number of distinct integers present on the board after 109 days have elapsed. Note:  Once a number is placed on the board, it will remain on it until the end. % stands for the modulo operation. For example, 14 % 3 is 2.    Example 1:  Input: n = 5 Output: 4 Explanation: Initially, 5 is present on the board.  The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.  After that day, 3 will be added to the board because 4 % 3 == 1.  At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.   Example 2:  Input: n = 3 Output: 2 Explanation:  Since 3 % 2 == 1, 2 will be added to the board.  After a billion days, the only two distinct numbers on the board are 2 and 3.     Constraints:  1 <= n <= 100  

	# Explanation
	Here's a breakdown of the solution and the code:

*   **High-Level Approach:**
    *   The core idea is to simulate the process until a fixed point is reached, i.e., no new numbers are added to the board. Since the number of days is extremely large (10<sup>9</sup>), direct simulation would be infeasible.
    *   We maintain a set to keep track of the distinct numbers currently on the board.
    *   We iterate, adding new numbers based on the modulo condition, until no further numbers are added in a given iteration.

*   **Complexity:**
    *   Runtime Complexity: O(n<sup>2</sup>) - In the worst case, we might need to iterate almost up to 'n' times (each adding a few elements), and the modulo operation takes O(n) each time.  Since the board size is bound by 'n', and the number of iterations are also bound by n
    *   Storage Complexity: O(n) - The maximum number of distinct elements on the board cannot exceed 'n'.

	
	# Code
	```python
	def distinct_integers(n: int) -> int:
    """
    Calculates the number of distinct integers present on the board after 10^9 days.

    Args:
        n: The initial integer on the board.

    Returns:
        The number of distinct integers present on the board.
    """

    board = {n}
    while True:
        new_numbers = set()
        for x in list(board):  # Iterate over a copy to avoid modification issues
            for i in range(1, n + 1):
                if x % i == 1:
                    new_numbers.add(i)

        original_board_size = len(board)
        board.update(new_numbers)
        if len(board) == original_board_size:
            break

    return len(board)
	```